If A Pure-breeding (homozygous) Black (dominant), Long-haired (recessi...

If a pure-breeding (homozygous) black (dominant), long-haired (recessive) cat is mated to a pure- breeding Siamese, short-haired cat, and one of their male offspring is mated to one of their female offspring, what is the chance of producing a Siamese colored, short-haired kitten?
MAKE A PUNNET SQUARE

Answer:

To create a Punnett square for this problem, we need to first determine the genotypes of the parent cats.

The black, long-haired cat is homozygous dominant for black fur, so its genotype is BB. It is also homozygous recessive for long hair, so its genotype is ll.

The Siamese cat is homozygous recessive for black fur, so its genotype is bb. It is also homozygous dominant for short hair, so its genotype is SS.

Using this information, we can create the following Punnett square for the first generation (F1) of offspring:

B B

l Bl Bl

All of the F1 offspring are black and long-haired, because the black fur allele is dominant and the long hair allele is recessive. Each offspring has the genotype Bll.

Next, we can create a Punnett square for the second generation (F2) of offspring, which occurs when one male and one female F1 offspring mate:

Bl Bl

Bll BBl BBl

In this generation, there is a 100% chance of producing black offspring, because both parents are heterozygous for black fur and therefore carry one dominant black allele each. However, there is a 50% chance of producing a Siamese-colored kitten, because each parent has a recessive allele for black fur. There is also a 100% chance of producing short-haired offspring, because both parents are homozygous dominant for short hair.

Therefore, the chance of producing a Siamese-colored, short-haired kitten is 50%, or 1/2.

If A Pure-breeding (homozygous) Black (dominant), Long-haired (recessi…

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